A 500g copper at 100celsius is placed in a 200g water at 25celsius which is contained in an aluminum calorimeter that has a mass of 400g. If the final temperature of the whole system is 30celsius, find the specific heat of the lead.
To find the specific heat of lead, we need to use the principle of conservation of energy.
First, we can calculate the energy gained or lost by each object using the equation:
Q = mcΔT
where Q is the heat energy gained or lost, m is the mass of the object, c is the specific heat of the object, and ΔT is the change in temperature.
For the copper block:
Q1 = 500g * c_copper * (30°C - 100°C)
For the water:
Q2 = 200g * c_water * (30°C - 25°C)
For the aluminum calorimeter:
Q3 = 400g * c_aluminum * (30°C - 25°C)
According to the principle of conservation of energy, the heat lost by one object is equal to the heat gained by the other objects in the system. Therefore:
Q1 + Q2 = Q3
Substituting the values we know:
500g * c_copper * (30°C - 100°C) + 200g * c_water * (30°C - 25°C) = 400g * c_aluminum * (30°C - 25°C)
Simplifying this equation will give us the specific heat of lead:
500g * c_copper * (-70°C) + 200g * c_water * 5°C = 400g * c_aluminum * 5°C
Now, we need the specific heat of aluminum, which is 0.897 J/g°C. We can substitute this value into the equation:
500g * c_copper * (-70°C) + 200g * c_water * 5°C = 400g * 0.897 J/g°C * 5°C
Now, we can solve this equation for the specific heat of copper (c_copper). By rearranging the equation:
c_copper = (400g * 0.897 J/g°C * 5°C - 200g * c_water * 5°C) / (500g * -70°C)
Substituting the known values gives the final answer for the specific heat of copper.